The photoelectric cut off voltage in a certain experiment is 1.5 V. The maximum kinetic energy of photoelectrons emitted is then

Light of wavelength 0.6 mm from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is 0.5 V. With light of wavelength 0.4 mm from a sodium lamp, the stopping potential is 1.5 V. With this data, the value of h/e is:

Photons absorbed in meter are converted to heat. A source emitting $n$ photons/s of frequency $v$ is used to convert $1\ kg$ of ice of ${0}^{o}c$ to water at ${0}^{o}C$. Then, the time taken for the conversion:

The energy od proton in a light of wavelength 6000 $\mathring { A } $ is :
$(h = 6.63 \times 10^{-34}$ $Js,$ $c = 3 \times 10^8 m/s)$

A photo-electric threshold wavelength of tungsten is $2300 \mathring{A}$ . The energy of electrons ejected from the surface, if ultra -violet light of wavelength $1800 \mathring{A}$ is incident on it, is 

Calculate the kinetic energy of the electron having wavelength 1 nm

A laser beam ($\iota = 633\ nm$)has an power of $3mW$. What will be the pressure exerted on a surface by this beam if the cross sectional area is $3\ mm^{2}$ (Assume perfect reflection and normal incidence)

A metalic surface is irradiated with monochr matic light of variable wavelength. Above wavelength of $ 5000 \dot { A }  $ , no photoelectrons a emitted from the surface. With an unknown wavelength, a stopping potential of 3V is nessary to eliminate the photo current . The Known wavelength is:

Two free protons are separated by a distance of $1\mathring A $. If they are released the kinetic energy of proton when it infinite separation is

The maximum energy of emitted photo electrons is measured by

If the work function of the metal is W and the frequency of the incident light is v , then there is no emission of photo-electrons if

A proton, an electron and an $\alpha$ particle is accelerated through the same potential difference enter a region of uniform magnetic field, moving at right angles to the magnetic field $\vec{B}$. The ratio of their kinetic energies is

Identify which of the following statement is true about the momentum of a photon?

What should be the velocity of an electron so that its momentum becomes equal to that of a photon of wavelength $5200\overset {\circ}{A}$?

Ultraviolet radiation of 6.2 eV falls on an aluminium surface (work function 4.2 eV). The kinetic energy (in joule) of the fastest electron emitted is :

Momentum of a photon having frequency $1.5\times 10^{13}Hz$?

A beam of light has two wavelengths $4971\mathring{A}$ and $6216\mathring{A}$ with a total intensity of $3.6\times { 10 }^{ -3 }W{ m }^{ -2 }$ equally distributed among the two wavelengths. The beam falls normally on an area of $1{cm}^{2}$ of a clean metallic surface of work function $2.3eV$. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photo electrons liberated in $2s$ approximately :

The threshold frequency for a metallic surface corresponds to an energy of $6.2eV$, and the stopping potential for a radiation incident on this surface $5V$. The incident radiation lies in.

When photon of the energy 3.8 eV falls on metallic surface of work function 2.8 eV, then the kinetic energy of emitted electrons are

An electron of mass $m$ and charge $e$ is accelerated from rest through a potential difference of $V$ volt in vacuum. Its final speed will be

A hydrogen-like atom is in a higher energy level of quantum number $6$. The excited atom make a transition to first excited state by emitting photons of total energy $27.2\ eV$. The atom from the same excited state make a transition to the second excited state by successively emitting two photons. If the energy of one photon is $4.25\ eV$, find the energy of other photon.

Calculate the number of photons emitted per seconds from a monochromatic light source of $40\ W$, giving out light of wavelength $5000\ \mathring {A}$.

In a photoelectric cell, illuminated with a certain radiation, the minimum negative anode of potential with respect to emitting metal required to stop the electron is $2 V.$ the  minimum KE of the photoelectrons is 

The difference in the electron energies associated wiwth the two state of an atom is $4 eV$. if $\frac { h }{ e } =4\times { 10 }^{ -5 }{ JsC }^{ -1 }$, the wavelength of the photon emitted as a result of the above transition will be

When a hydrogen atom emits a photon of energy $12.09eV$,it's orbit's angular momentum changes by (where $h$ is Planck's constant) ?

Assuming photo-emission to take place, the factor by which the maximum velocity of the emitted photo electrons changes when the wavelength of the incident radiation is increased four times, is (assuming work function to be negligible in comparison to $hcl\lambda $)

Momentum of $\gamma-ray$ photon of energy $3\ keV$ in $kg-m/s$ will be

Radiational wave length $ \lambda $=124 nm falls on a metallic surface. Then the kinetic energy of the ejected photo electron(s) can be : (Given that threshold wavelength ($ \lambda _{0} $)=248 nm)

Total energy of $electron$ is more than energy of $photon$ if both are having $equal\ \lambda.$

Monochromatic light of wavelength 440 nm is produced. The power emitted by light is 18 mW, The number of photons emitted per second by light beam is :

An electron of stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be

Violet light is falling on a photosensitive material causing ejection of photoelectrons with maximum kinetic energy of $1$ eV. Red light falling on metal will cause emission of photoelectrons with maximum kinetic energy (approximately) equal to

When an electron de-exited back from ${\left( {n + 1} \right)^{th}}$ state to ${n^{th}}$ state in a hydrogen like atoms, wavelength of radiations emitted is ${\lambda _1}\left( {n >  > 1} \right)$. In the same atom de-broglies wavelength associated with an electron in $nth$ state is ${\lambda _2}$. Then $\frac{{{\lambda _1}}}{{{\lambda _2}}}$ is proportional to 

Approximately, the temperature corresponding to $1 eV$ translational kinetic energy of molecule is

The work function of a metal is $3.3\times{  10}^{  -19}J$. The maximum wavelength of the photons required to eject electron from the metal is:

A metal plate of area $1\times { 10 }^{ -4 }{ m }^{ 2 }$ is  illuminated by a radiation of intensity 16 m $W/{ m }^{ 2 }.$ The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be :
$\left[ { 1eV=1.6\times 10^{ 19 }J } \right] $

The energy of a hydrogen-like atom (or ion ) in its ground state is - 122.4 eV. It may be :

When radiations of wavelength 3000 are incident on a photosensitive surface, the kinetic energy of electrons is 2.5 eV. The stopping potential for 1500 will he,

A photon of energy 12.75 eV is completely absorbed by a hydrogen atom initially in the ground state. The quantum number of the excited state is:

If an $\alpha$-particle is accelerated with the potential V, the energy of $\alpha$-particle will be 

If an $\alpha$-particle and a proton are accelerated from rest by a potential difference of 1 megavolt then the ratio of their kinetic energy will be

A photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\cfrac{\lambda}{2}$. If the maximum kinetic energy of the emitted photoelectrons in the second case is $3$ times that in the first case, the work function of the surface of the material is  ($h=$ Planks constant, $c=$ speed of light)

In photoelectric effect for silver threshold is $\lambda _0 = 3250 \times 10^{-10} m$. If U.V of $\lambda = 2536 \times 10^{-10}$ is incident then velocity of electron from will be 

Find the maximum $KE$ of photoelectrons emitted from the surface of lithium$(\phi=2.39 eV)$ when exposed with $\displaystyle E=E _{0}(1+\cos 6\times10^{14}t)\cos 3.6\times 10^{15}t$

Two separate monochromatic light beams A and B of the same intensity are falling normally on a unit area of a metallic surface. Their wavelengths are $\lambda _A$ and $\lambda _B$, respectively. Assuming that all the incident light is used in ejecting the photoelectrons, the ratio of the number of photoelectrons from beam A to that from B is

The momentum of a photon of energy 1MeV, in kg/m/s, will be :

When light is made incident on a surface, then photoelectrons are emitted from it. The kinetic energy of photoelectrons

The work function of aluminum is 4.2eV. Light of wavelength $\displaystyle 2000\mathring {A} $ is incident on it. The maximum energy of emitted electrons will be

The work function of aluminum is 4.2eV. Light of wavelength $\displaystyle 2000\mathring {A} $ is incident on it. The minimum energy of emitted electrons will be 

The work function of a metal is X eV. When light of energy 2X is made incident on it then the maximum kinetic energy of emitted photoelectron will be 

A 14 eV energy photon ionises a hydrogen atom in its lowest energy level. The kinetic energy of the electron ejected from the atom will be :

The threshold wavelength of a photosensitive metal is $662.5\ nm$. If this metal is irradiated with a radiation of wavelength $331.3\ nm$, find the maximum kinetic energy of the photoelectrons.

The wavelength of a photon is $5000\mathring {A} $.Its momentum will be:

Momentum of $ \gamma -ray $ photon of energy 3 KeV in kg-m/s will be 

What is the momentum of a photon having frequency $1.5 \times 10^{13} Hz $ 

The frequency and the intensity of a beam of light falling on the surface of a photoelectric material are increased by a factor of two. This will :

The frequency of incident light falling on a photosensitive metal plate is doubled, the KE of the emitted photoelectrons is

When orange light falls on a photo sensitive surface the photocurrent begins to flow.Velocity of emitted electrons will be more when surface is hit by

Assertion (A) : For a fixed incident photon energy, photoelectrons have a wide range of energies ranging from zero to the maximum value $K _{max}$
Reason (R) : Initially, the electrons in the metal are at different energy level.

When light falls on a photosensitive surface, electrons are emitted from the surface. The kinetic energy of these electrons does not depend on the:

The photoelectrons emitted from a metal surface are such that their velocity

The photoelectric threshold of Tungsten is 2300$ \mathring {A }$. The energy of the electrons ejected from the surface by ultraviolet light of wavelength 18000 $ \mathring {A }$ is 

When light is incident on a metal surface the maximum kinetic energy of emitted electrons

Ultraviolet radiation of 6.2eV falls on an aluminium surface (work function 4.2eV). The kinetic energy in joule of the faster electron emitted is approximately

The ratio of de-Broglie wavelengths of proton and $\alpha$-particle having same kinetic energy is

$K _{1} $and $K _{2}$ are the maximum kinetic energies of the photoelectrons emitted when light of wavelength $\lambda _{1} $ and $\lambda _{2} $  respectively are incident on a metallic surface. If $\lambda _{1}= $3$\lambda _{2} $  then

The work function of a metal is $1.6\times 10^{-19}$J. When the metal surface is illuminated by the light of wavelength 6400 $A^{o}$, then the maximum kinetic energy of emitted photoelectrons will be ($h = 6.4 \times 10^{-34} Js$)

The work function of a metal is 4.6eV. The wavelength of incident light required to emit photo-electrons of zero energy from its surface, will be

The photoelectric work function of a metal surface is 2eV. When light of frequency $1.5 \times10^{15}$ Hz is incident on it, maximum kinetic energy of the photo-electrons, approximately is :

Work function of a metal is 3.0eV. It is illuminated by a light of wavelength $3 \times 10^{-7}$m. Then the maximum energy of the electron is.

The energy of the incident photon is 12.38 eV, while the energy of the scattered photon is 9.4 eV. The K.E. of the recoil electron is nearly

Light of wavelength 5000 $A^{o}$ falls on a sensitive plate with photoelectric work function 1.9eV. The maximum kinetic energy of the photoelectrons emitted will be

When light of wavelength 2480 $A^{0}$ is incident on a metal surface electrons are emitted with a maximum KE of 2 eV. The maximum KE of photo-electrons, if light of wavelength 1240 $A^{0}$ is incident on the same surface would be

The cut-off voltage in a photoelectric experiment is 3V. Then the maximum KE of photo-electrons emitted is

If the frequency of light incident on a photosensitive metal plate is doubled, then the KE of photoelectrons will be

The photo-electric threshold wavelength for a metal is $5000 A^{0}$. Light of wavelength $4000 A^{0}$ is incident on it. The maximum KE of photo-electrons emitted is [given $hc= 2 \times 10^{-25} Jm$]

An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.40$A^{0}$. Then the maximum energy of the photon of the emitted radiation. 

The work function of a certian metal is 2.3 eV .If light of wave number $2\times10^{6}m^{-1}$ falls on it,the kinetic energies of fastest and slowest ejected electorn will be respectively:

The ratio of momentum  of an electron and an $\alpha$-particle which are accelerated from rest by a potential difference of 100 V is

Which one of the following branch of physics deals the impossibility of making simultaneous, arbitrarily precise measurements of the momentum and the position of an electron.

In photoelectric effect, initially when energy of electrons emitted is $E _{0}$, de-Broglie wavelength associated with them is $\lambda _{0}$. Now, energy is doubled then associated de-Broglie wavelength $\lambda^{'}$ is

When the momentum of a photon is changed by an amount $p'$ then the corresponding change in the de-Broglie wavelength is found to be $0.20$%. Then, the original momentum of the photon was

In a photo electric effect experiment, the maximum kinetic energy of the emitted electrons is $1eV$ for incoming radiation of frequency $v _{0}$ and $3eV$ for incoming radiation of frequency $3v _{0}/2$. What is the maximum kinetic energy of the electrons emitted for incoming radiations of frequency $9v _{0}/4$?

Light of wavelength $\lambda$ strikes a photo sensitive surface and electrons are ejected with kinetic energy $E$. If the kinetic energy is to be increased to $2E$, then the wavelength must be changed to $\lambda'$, where :

A photo-cell is illuminated by a source of light, which is placed at a distance $d$ from the cell. If the distance become $d/2$, then number of electrons emitted per second will be : 

The formula for kinetic mass of photon is:
where $h$ is Planck's constant and $v,\lambda, c$ are frequency, wavelength and speed of photon respectively.

If momentum of a photon of an electromagnetic radiation is $3.3\times 10^{-29}$ kg m/sec, then frequency of associated wave is:

Assume that a neutron breaks into a proton and an electron. The energy released suring this process is (mass of neutron $= 1.6725 \times 10^{-27} kg,$
mass od proton $= 1.6725 \times 10^{-27} kg,$ mass of electron$= 9 \times 10^{-31}kg)$

A monochromatic source of light is placed at a large distance $d$ from a metal surface. Photoelectrons are ejected at rate $n$, the kinetic energy being $E$. If the source is brought nearer to distance $d/2$, the rate and kinetic energy per photoelectron become nearly :

The work function of a certain metal is $3.31 \, \times \,  10^{-19} J.$ Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength 5000 A is
$(Given\, h  = 6.62\times10^{-34} J-s, \, c= 3\times10^{-8} \, ms^{-1},\, e=  1.6 \times10^{-19} \, C)$